A random sample of n observations is selected from a normal population to test the null hypothesis that µ=10.Specify the rejection region for each of the following combinations of \(Ha,\alpha ,\) and n:

a.\(Ha:\)µ\( \ne 10;\alpha = .05.;n = 14\)

b.\(Ha:\)µ\( > 10;\alpha = .01;n = 24\)\(\)

c.\(Ha:\)µ\( > 10;\alpha = .10;n = 9\)

d.\(Ha:\)µ <\(10:\alpha = .01;n = 12\)

e.\(Ha:\)µ\( \ne 10;\alpha = .10;n = 20\)

f. \(Ha:\)µ<\(10;\alpha = .05;n = 4\)

Referring to pages 405,406, 407, 414.Given that µ=10. We must define the rejection region for the following combination:\(Ha,\alpha ,n\)

a)

Test hypotheses are as follows.

\(H0:\)µ\( = 10\)

\(Ha:\)µ\( \ne 10\)

\(\begin{aligned}n &= 14\\df &= n - 1\\df &= 13\end{aligned}\)

Referring to a t distribution table, the two-tailed test means \(\frac{\alpha }{2}\) because the total \(\alpha \) is 0.05. Keeping the following in mind:

The significance level is 0.05

The critical value is as follows.

\(\begin{aligned}tc &= TINV\left( {0.05,13} \right)\\tc &= 2.160\\\end{aligned}\)

Reject \(H0\) if \(t < - 2.160\) or \(t > 2.160\)

b)

Test hypothesis are as follows

\(H0 = 10\)

\(Ha:\)µ>10

\(\begin{aligned}\alpha & = 0.01\\n &= 24\end{aligned}\)

Because of the alternative hypothesis, this is a two-tailed test.

\(\begin{aligned}df &= n - 1\\df &= 24 - 1\\df &= 23\end{aligned}\)

Referring to a t distribution table, the two-tailed test means alpha\(\frac{\alpha }{2}\) because the total\(\alpha \) is 0.01, one tail indicates that we should keep the\(\alpha \)as. Keeping the following in mind:

The significance level is\(0.01\)

The critical value is as follows.

\(\begin{aligned}tc &= T.INV\left( {1 - 0.01,23} \right)\\tc &= 2.500\\\end{aligned}\)

Reject \(H0\) if \(t > 2.500\)

c)

Test hypothesis are as follows

\(H0 = 10\)

\(Ha:\)µ>.10

\(n = 9\)

Because of the alternative hypothesis, this is a two-tailed test.

\(\begin{aligned}df &= n - 1\\df &= 9 - 1\\df &= 8\end{aligned}\)

Referring to a t distribution table, the two-tailed test means alpha\(\frac{\alpha }{2}\) because the total\(\alpha \) is 0.10, one tail indicates that we should keep the\(\alpha \)as. Keeping the following in mind:

The significance level is\(0.10\)

The critical value is as follows.

\(\begin{aligned}tc &= T.INV\left( { - 0.10,8} \right) &= 1.372\\tc &= 1.372\end{aligned}\)

Reject\(H0\)if\(t > 1.372\)

d) Test hypothesis are as follows

\(H0:\)µ\( = 10\)

\(Ha:\)µ<10

\(n = 12\)

\(\begin{aligned}df &= n - 1\\df &= 12 - 1\\df &= 11\end{aligned}\)

Referring to a t distribution table, the two-tailed test means alpha\(\frac{\alpha }{2}\) because the total\(\alpha \) is 0.01, one tail indicates that we should keep the\(\alpha \)as. Keeping the following in mind:

The significance level is\(0.01\)

The critical value is as follows.

\(\begin{aligned}tc &= T.INV\left( {0.01,11} \right)\\tc &= 2.821\end{aligned}\)

Reject \(H0\) if \(t < - 2.281\)

e) Test hypothesis are as follows

\(H0:\)µ\( = 10\)

\(Ha:\)µ\( \ne 10\)

\(n = 20\)

\(\begin{aligned}df &= n - 1\\df &= 20 - 1\\df &= 19\end{aligned}\)

Referring to a t distribution table, the two-tailed test means alpha \(\frac{\alpha }{2}\) because the total \(\alpha \) is 0.10, one tail indicates that we should keep the \(\alpha \) as. Keeping the following in mind:

The critical value is as follows.

\(\begin{aligned}tc &= T.INV.2T\left( {0.10,19} \right)\\tc &= 1.725\end{aligned}\)

Reject \(H0\) if t<-1.725 or t >1.725

f) Test hypothesis are as follows

\(H0:\)µ\( = 10\)

\(Ha:\)µ<10

\(n = 4\)

\(\begin{aligned}df &= n - 1\\df &= 4 - 1\\df &= 3\end{aligned}\)

Referring to a t distribution table, the two-tailed test means \(\frac{\alpha }{2}\) because the total \(\alpha \) is 0.05. Keeping the following in mind:

The significance level is 0.05

The critical value is as follows.

\(\begin{aligned}tc &= T.INV\left( {0.01,3} \right)\\tc &= - 2.353\\\end{aligned}\)

Reject \(H0\)if t<-2.353